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I cannot answer this question for my life!

  1. Sep 16, 2010 #1
    1. The problem statement, all variables and given/known data
    A falling stone takes .28s to travel past a window 2.2m tall. How far above the window was it released


    2. Relevant equations
    vf=vi+at
    d=vit+(1/2)at^2
    vf^2=vi^2+2ad


    3. The attempt at a solution
    vf^2=(6.5)+2(9.81 x 2.2)

    2.2=(x)(.28s)+(1/2)(9.81)(.28^2)
     
  2. jcsd
  3. Sep 16, 2010 #2

    kuruman

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    Hi High_Voltage and welcome to PF. you have the right relevant equations, but you are not applying them correctly. Start with d=vi*t+(1/2)a*t^2
    You know d=2.2 m, t=0.28 s and a = 9.8 m/s2 (assuming that "down" is positive). You don't know what vi is but you can find it from this equation. It represents the speed of the stone when it first appears at the top of the window. Once you have a number for vi you need to ask and answer another question: How far must a stone (that is released from rest) have to drop so that is reaches this speed vi? That's the answer you are looking for.
     
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